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Coherent State Confusion

  1. Mar 13, 2014 #1
    So a coherent state in quantum mechanics is "the most classical" quantum state (A Gaussian wave packet), which satisfies the Heisenberg uncertainty relation with an equality. This allows the wave packet to travel in space in a more localized fashion (like a classical particle) because its uncertainty doesn't "float away" in time.

    I have two questions.

    The first question is, I have heard people say a laser is an example of a coherent state. Why is this? I know in the classical explanation, it is because all the waves inside the laser cavity are in phase and localized to a beam, this is said to be very coherent. But is a laser also an example of a quantum coherent state? And why?

    My second question is, how would one go about MAKING these coherent state photons experimentally? I want to have some physical backing behind my understanding. Would an extremely attenuated laser pulse (so it is down to just a few photons) be an example of photons in a coherent state?

    If anyone could help me understand this, it would be great!
     
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  3. Mar 13, 2014 #2

    bhobba

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    Its being used in two different senses.

    For laser coherency it means the photons have exactly the same polarisation - for normal light the photons have random polarisation.

    In the case of wave-packets its a state that has the properties of the least spread in position and momentum allowed by the Heisenberg uncertainly principle. But they actually aren't stable - initially they start as a clump but they tend to 'spread' and exactly how our stable, non spreading, classical everyday world emerges is an interesting question - but that would take us too far afield.

    What you likely have been confused by is the Wikipedia article:
    http://en.wikipedia.org/wiki/Coherent_states#The_wavefunction_of_a_coherent_state

    Yes - it is correct - at a deep technical level all these things have a connection - but unless you are into the technicalities then its not the best way to view it at the layman or beginning student level.

    Thanks
    Bill
     
    Last edited: Mar 13, 2014
  4. Mar 13, 2014 #3
    Thank you for your reply!

    So basically a laser would be consider coherent in the classical sense, but not wouldn't represent a quantum coherent state?

    My next question is how would one go about making a coherent state experimentally?
     
  5. Mar 13, 2014 #4

    bhobba

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    I don't know what you mean by coherent in the classical sense. Photon polarisation is a quantum concept.

    Mate I have zero idea.

    Bose-Eisten condensates may be along the lines of what you are after:
    http://en.wikipedia.org/wiki/Bose–Einstein_condensate

    But its an interesting thing - even though most of the particles of the condensate are in the lowest energy state and in that sense coherent QM itself predicts all of them cant be.

    So maybe complete coherence is a chimera?

    Thanks
    Bill
     
  6. Mar 14, 2014 #5

    DrDu

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    Bill, I am not quite d'accord with some of your statements. Coherence is not about polarization. Obviously you can have perfectly linearly or circularly polarized beam of light with little coherence. Classically, there are two kinds of coherence, spatial and temporal coherence. Spatial coherence is limited by the light field being a random mixture of Fourier components with slightly different k vector, temporal coherence by light wave being a mixture of states with different frequency ω.
    In the case of a laser, there are quantum coherence effects which don't have a classical analog, e.g. higher order correlations between multi photon events. Clearly they don't have a classical analog as there are no photons classically.
    A light wave which has full coherence in all orders is a coherent state.
    In the case of light, coherent states are stable and don't spread out, at least in vacuo, due to the linear dispersion ##\omega =c k##. This is in contrast to the behaviour of, say, a free electron wavepacket.
    I think a good starting point is the Nobel lecture of Glauber:
    http://www.nobelprize.org/nobel_prizes/physics/laureates/2005/glauber-lecture.pdf
     
  7. Mar 14, 2014 #6

    kith

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    You cannot make a "coherent state photon" because coherent states are states with an indefinite number of photons. They are eigenstates of the annihilation operator so they have the remarkable property that the annihilation of a photon leaves the state unchanged.

    The experimentalist's answer would be because it has the properties of a coherent state. ;-) Personally, I'd like to see a simple model of a laser where the time evolution yields a coherent state for the light field. Open quantum dynamics people have probably done this, but I don't know it.

    Classically, laser light is a wave with both definite amplitude and phase. Quantum mechanically, we have an uncertainty relation between photon number (which is related to the amplitude) and phase, so we need a compromise. Like in the case of the harmonic oscillator with position and momentum, there is a set of states of the light field with minimal uncertainty in photon number and phase: the coherent states. /edit: Wikipedia has a nice illustration of this.

    But note that this is a handwavy argumentation because the phase operator and the uncertainty relation including it are problematic concepts.
     
    Last edited: Mar 14, 2014
  8. Mar 14, 2014 #7

    Cthugha

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    As has been pointed out, coherent states are eigenstates of the annihilation operator. This already points out the analogy to classical states of the light field as a measurement does not change the state of the field (just like a measurement in classical physics).

    Whether or not, you gain additional knowledge about a light field when performing a measurement depends on two things:
    a) Noise. If you have a very noisy photon number distribution, the probability to detect a photon will be high when the instantaneous intensity is high and it will be low when the instantaneous intensity is low. So a noisy light field gives you some information.
    b) Photon annihilation. The detection of a photon destroys it. So you now, that you just reduced the photon number. For example, if you have just one photon present and detect it, you will know that there will be no second photon detection event.

    So the first effect tends to increase the conditional probability to detect a second photon, while the second effect tends to decrease the conditional probability to detect a second photon. This means that there must be some intermediate level of photon number distribution noise at which the conditional photon detection probability is neither increased nor decreased.

    If you do the math, you will find that this is exactly the case for a process which has Poissonian statistics, which is realized for stimulated emission like in a laser.
     
  9. Mar 14, 2014 #8

    kith

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    Is noise simply another word for uncertainty? So classical noise is related to the purity of the state and quantum noise to a pure state not being an eigenstate of the observable of interest?

    I don't really get why you say that noise is the reason for this. If the instantaneous intensity is high, the average number of photons present is high, so we should have a high probability of detecting one and vice versa. What am I missing?
     
  10. Mar 14, 2014 #9

    Cthugha

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    If you want to think about it that way, you can consider photon number noise a sphoton number uncertainty.

    What makes you think that there is a connection between instantaneous intensity (not mean intensity) and the average number of photons present?

    Consider two different fields, both with an initial average photon number of 10 photons.

    a) There are initially exactly 10 photons. In that case detecting a photon will reduce your photon number to 9. Another detection will reduce it to 8 and so on.

    b) A field that has 0 photons 50% of the time and 20 photons 50% of the time. If you detect a photon, you will have 19 photons left and thus you have a drastically increased probability to detect a second photon compared to case a), although you have the same mean photon number.

    The broader your distribution is, the higher will be your conditional probability to detect a second photon.
     
  11. Mar 14, 2014 #10

    kith

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    Isn't the instantaneous intensity defined as I(t) = const. * <N(t)>? Now that I think about it, this doesn't seem sensible because the average photon number should be constant. What is the definition of the instantaneous intensity then?
     
  12. Mar 14, 2014 #11

    Cthugha

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    Oh, I see.

    The light fields you typically encounter are not photon number eigenstates and thus have a photon number that is not fixed over time.

    For a good description of the light field in terms of the photon number, you therefore need to things: The mean photon number and a distribution function that tells you the probability of having a certain photon number present at some instant (at fixed mean photon number). The typical timescale over which the photon number fluctuates is given by the coherence time of your light field.

    The distribution functions you encounter in reality are somewhat surprising. For a laser you get a Poissonian distribution, which is still pretty centered around the mean value. But for, say, light from the sun or an incandescent light bulb, the distribution will be a Bose-Einstein distribution. That has the interesting consequence that 0 is the most probable photon number at any instant and quite some of the intensity you see can be attributed to really high photon numbers that occur with pretty small probability. However, as these fluctuations occur on a femtosecond to nanosecond scale, they are way to fast to see them with your eyes.
     
  13. Mar 14, 2014 #12

    naima

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    Is because expectation value E = variance V?
    How is p(n knowing 1 photon) linked to E and V?
     
    Last edited: Mar 14, 2014
  14. Mar 15, 2014 #13
    I am not sure if I understand 100%, so help me out on this one. From what I understand from your post, a Poissonian distribution is the case where the conditional photon detection probability is neither increased or decreased. This distribution is realized in a stimulated emission such as a laser.

    So my first thought is, "I know coherent state photons exhibit a Poissonian distribution, thus a stimulated emission from a laser emits quantum coherent states because it also exhibits this distribution."

    Would this be correct?
     
  15. Mar 16, 2014 #14

    naima

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    before detecting one photon the Poissonian probability for k photons is [itex]\lambda^k e^{-\lambda}/k![/itex]
    we have [itex]p_0 + p_1 + p_2 + ... = 1[/itex]
    once we have found one photon we get other probabilities [itex]p'_0 p'_1 p'_2[/itex] .with
    [itex]p'_k.= p_{k+1}/(1 - p_0)[/itex]
    I do not see that conditional probabilities remain constant
     
  16. Mar 17, 2014 #15

    Cthugha

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    Ok, the thorough definition of coherence is given by a correlation function that factorizes in all orders. However, this is not really helpful for getting a basic understanding. One can loosely interpret the normalized nth-order correlation function as the conditional probability of detecting the nth photon after you have already detected n-1 photons. It is normalized with respect to the case of statistically independent photons.

    Now one can show that the Poissonian distribution has several features which match what you expect for the coherent state. The Poissonian distribution is the distribution for statistically independent events. Therefore it is not surprising that it gives you a normalized correlation function that factorizes. Also, if you take the coherent state and apply the photon annihilation operator to it, you will find that the average photon number does not change. This is also matching the statistical independence requirement.

    So basically, yes, it is somewhat like the duck test: You expect to have some properties and stimulated emission shows these properties. There is some more to it, but that is somewhat deep. The lecture by Glauber which was linked earlier really is a very good start to introduce yourself to the topic.

    The one point you need to keep in mind is the timescale. The correlation functions needs to factorize at timescales shorter than the coherence time of the light in order for the state to be a coherent one. On long timescales any light field will look like a Poissonian one. This is because photons from a single mode are only indistinguishable within their coherence time.

    That is not really a conditional probability. You would have to calculate the photon number distribution in the case that one photon was already detected (so p_0 is out, for example). Let me roughly sketch the math for the equal-time second-order correlation function. It is defined as:
    [tex]g^{(2)}(0)=\frac{\langle n (n-1) \rangle}{\langle n \rangle^2}[/tex].

    Here, the n is the instantaneous photon number. This quantity gives you the normalized probability to detect a second photon after you already detected one (the n-1-term comes from normal ordering and represents the destruction of a photon by the first detection process). It is the relative probability compared to the case of a light field with the same mean photon number, but photon detection events which are statistically completely independent. In that case the expected number of detections would be [itex]\langle n \rangle^2[/itex] as shown in the denominator.

    You can express the instantaneous photon number n as the combination of the mean value and some fluctuation around it: [itex]n=\langle n \rangle +\delta n.[/itex]

    That gives you:
    [tex]g^{(2)}(0)=\frac{\langle \langle n \rangle^2 + 2 \langle n \rangle \delta n - \langle n \rangle + \delta n^2 -\delta n \rangle}{\langle \langle n \rangle +(\delta n) \rangle^2}[/tex]

    The expectation value of the fluctuation is of course 0, so all terms linear in the fluctuation will vanish. That leaves you with:
    [tex]g^{(2)}(0)=\frac{\langle n \rangle^2 - \langle n \rangle + \langle (\delta n)^2 \rangle}{\langle n \rangle^2}.[/tex]
    Or in short:
    [tex]g^{(2)}(0)= 1 -\frac{1}{\langle n \rangle}+\frac{\langle (\delta )n^2 \rangle}{\langle n \rangle^2}.[/tex]

    A value of 1 gives you an unchanged conditional probability. Values above 1 describe a tendency for photons to bunch (like for sunlight). Values below 1 represent a tendency for photons to avoid each other (like for single photon sources).

    Now the first term is always 1. The second term is always negative and represents the destruction of a photon in the first detection process. The third term is always positive and describes a tendency to bunch in terms of the fluctuations present in the light field. The central quantity [itex]\langle (\delta n)^2 \rangle[/itex] is the variance of the photon number distribution.

    So in order to get a value of 1 and an unchanged conditional probability, you need the second and the third term to cancel exactly. As one can easily see, this will happen when the variance of the distribution is exactly equal to the mean value of the photon number. This is exactly the case for the Poissonian distribution. It is also the smallest relative variance you can get from a classical light field. All light fields showing a smaller relative variance are inherently non-classical.
     
  17. Mar 17, 2014 #16

    DrDu

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    I am not quite sure what you are referring to. Evidently, measuring photon number will leave you in an eigenstate of photon number and you will no longer have a Poissonian distribution. However, a coherent state is an eigenstate of the anihilation operator, which is not the same as being an eigenstate of the particle number operator.
     
  18. Mar 17, 2014 #17

    naima

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    I had not understood what Chtugha wanted to say. I will study his new accurate answer.
     
  19. Mar 19, 2014 #18

    naima

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    Cthugha first talked about super Poissonian light. The variance is greater than mean. It is the case of thermal light where "bunching" appears. At the opposite is sub Poissonian light with variance less than mean and with "antibunching". (it seems that the relation with bunching/antibunching is not so simple) And between coherent states with equallity.Most lasers are like that.
    I found this paper.
    It shows that lasers may be sub Poissonian. I think that the light of these lasers remains coherent and that the word coherence is not only for coherent states.
     
  20. Mar 19, 2014 #19

    Cthugha

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    You can get several deviations from ideal coherence for some kinds of lasers, for example single or few atom lasers, single or few quantum dot lasers, lasers with non-standard pump sources and so on. These lasers are not coherent in the quantum optics sense as coherence is defined in terms of factorizing correlation functions.

    These lasers may still be pretty coherent in the classical sense, though (having long field coherence lengths and times). Note that these two concepts of coherence are entirely different. However, only the first one is a yes/no criterion. You can have long or short classical coherence times, but there is no well defined line between coherent and incoherent for classical field coherence.
     
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